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Upon mixing \(45.0 \mathrm{~mL}\) of \(0.25 M\) lead nitrate solution with \(25 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble. \(\quad\) [I.I.T. 1993] [Hint : The reaction is: \(3 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3} \longrightarrow 3 \mathrm{PbSO}_{4}+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) 3 moles \(\quad\) I mole 3 moles 2 moles No. of moles of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}=45 \times 10^{-3} \times 0.25\) \(=11.25 \times 10^{-3}\) mole No. of moles of \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}=25 \times 10^{-3} \times 0.1=2.5 \times 10^{-3}\) mole Thus, \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) has limiting concentration. It shall be consumed fully and the number of moles of lead sulphate produced will be \(=3 \times 2.5 \times 10^{-3}=7.5 \times 10^{-3}\) mole No. of moles of lead nitrate left \(=11.25 \times 10^{-3}-7.5 \times 10^{-3}\) \(=3.75 \times 10^{-3}\) mole Total volume \(=(45.0+25.0)=70 \mathrm{~mL}\) or \(70 \times 10^{-3}\) litre $$ \text { Molarity }=\frac{3.75 \times 10^{-3}}{70 \times 10^{-3}}=0.0536 M $$ No. of moles of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) formed \(=2 \times 2.5 \times 10^{-3}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) will be present in solution in ionic form. Thus, $$ \left[\mathrm{Pb}^{2+}\right]=0.0536 M $$ $$ \begin{aligned} \left[\mathrm{Cr}^{3+}\right] &=0.0714 M \\ \left[\mathrm{NO}_{3}^{-}\right] &=(2 \times 0.0536)+(3 \times 0.0714) \\ &=0.3214 \mathrm{M}] \end{aligned} $$

Step by step solution

01

Calculate Moles of Reactants

First, calculate the number of moles of each reactant using the formula: \( ext{moles} = ext{concentration} \times ext{volume} \). For lead nitrate: \( 45 \times 10^{-3} \times 0.25 = 11.25 \times 10^{-3} \) moles. For chromic sulphate: \( 25 \times 10^{-3} \times 0.1 = 2.5 \times 10^{-3} \) moles.

02

Identify the Limiting Reagent

Refer to the reaction equation: \( 3 \text{Pb(NO}_3)_2 + \text{Cr}_2(\text{SO}_4)_3 \rightarrow 3 \text{PbSO}_4 + 2 \text{Cr(NO}_3)_3 \). It shows 3 moles of Pb(NO鈧�)鈧� react with 1 mole of Cr鈧�(SO鈧�)鈧�. Calculate required moles: Pb(NO鈧�)鈧� required = \( 3 \times 2.5 \times 10^{-3} = 7.5 \times 10^{-3} \) moles. Since 11.25x10^{-3} moles of Pb(NO鈧�)鈧� are available, Cr鈧�(SO鈧�)鈧� is the limiting reagent.

03

Calculate Moles of Product Formed

Since Cr鈧�(SO鈧�)鈧� is limiting, all of it will react. According to stoichiometry (3:1 ratio), the moles of PbSO鈧� produced are: \( 3 \times 2.5 \times 10^{-3} = 7.5 \times 10^{-3} \) moles.

04

Determine Remaining Moles

Lead nitrate reacts with Cr鈧�(SO鈧�)鈧�, consuming \( 7.5 \times 10^{-3} \) moles of Pb(NO鈧�)鈧�. Remaining moles of Pb(NO鈧�)鈧� is \( 11.25 \times 10^{-3} - 7.5 \times 10^{-3} = 3.75 \times 10^{-3} \) moles.

05

Calculate Total Volume

Add the volumes of the solutions: \( 45.0 \text{ mL} + 25.0 \text{ mL} = 70 \text{ mL} \) or \( 70 \times 10^{-3} \text{ L} \).

06

Calculate Molar Concentration of Species

The concentration of unreacted lead nitrate, Pb(NO鈧�)鈧�, is \( \frac{3.75 \times 10^{-3}}{70 \times 10^{-3}} = 0.0536 \text{ M} \). For Cr(NO鈧�)鈧� produced: \( 2 \times 2.5 \times 10^{-3} = 5.0 \times 10^{-3} \) moles, concentration is \( \frac{5.0 \times 10^{-3}}{70 \times 10^{-3}} = 0.0714 \text{ M} \).

07

Calculate Ionic Concentrations

Pb(NO鈧�)鈧� disassociates into Pb虏鈦� and NO鈧冣伝 ions, and Cr(NO鈧�)鈧� disassociates into Cr鲁鈦� and NO鈧冣伝 ions. \([\text{Pb}^{2+}] = 0.0536\text{ M}\). For NO鈧冣伝, arising from both sources: \( 2 \times 0.0536 + 3 \times 0.0714 = 0.3214 \text{ M} \). Thus, \([\text{Cr}^{3+}] = 0.0714\text{ M}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reagent

In chemical reactions, the limiting reagent is the substance that is completely consumed first, limiting the amount of product formed. In our exercise, we have a reaction between lead nitrate and chromic sulphate. We calculated the moles for each:

• Lead nitrate: 11.25 x 10^-3 moles
• Chromic sulphate: 2.5 x 10^-3 moles

The balanced chemical equation, \(3 \text{Pb(NO}_3)_2 + \text{Cr}_2(\text{SO}_4)_3 \rightarrow 3 \text{PbSO}_4 + 2 \text{Cr(NO}_3)_3\) shows us that 3 moles of lead nitrate react with 1 mole of chromic sulphate. Therefore, we need \(3 \times 2.5 \times 10^{-3} = 7.5 \times 10^{-3}\) moles of lead nitrate to fully react with chromic sulphate. Since we have more than enough lead nitrate, chromic sulphate is our limiting reagent. It tells us the maximum amount of product that can be formed, which in this case is 7.5 x 10^-3 moles of lead sulphate.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions react to form an insoluble solid, or precipitate. This exercise involves the formation of lead sulphate as a precipitate. Here's how it works:

• Lead nitrate solution, \(\text{Pb(NO}_3)_2\), and chromic sulphate solution, \(\text{Cr}_2(\text{SO}_4)_3\), are mixed.
• In the reaction, lead ions ([\text{Pb}^{2+}]) from lead nitrate combine with the sulphate ions ([\text{SO}_4^{2-}]) from chromic sulphate to form lead sulphate (\text{PbSO}_4), which is not soluble in water.

The solid lead sulphate forms as a precipitate in the solution. This exemplifies a classic precipitation reaction, where ionic components in aqueous solutions reassemble into a new compound that has low solubility in the existing solvent.

Molar Concentration

Molar concentration, or molarity, measures the concentration of a solute in a solution, defined as the number of moles of solute per litre of solution. In our problem, after the reaction, we're interested in the molar concentration of the remaining species.

• First, note the total volume of the solution: 70 mL or \(70 \times 10^{-3}\) L.
• The concentration of remaining lead nitrate (\text{Pb(NO}_3)_2) is calculated as \(\frac{3.75 \times 10^{-3}}{70 \times 10^{-3}} = 0.0536\, \text{M}\).
• The concentration of chromic nitrate (\text{Cr(NO}_3)_3) formed is \(\frac{5.0 \times 10^{-3}}{70 \times 10^{-3}} = 0.0714\, \text{M}\).

By calculating these concentrations, we understand the distribution of ions in the final solution, specifically \(\text{Pb}^{2+}, \text{Cr}^{3+},\) and \(\text{NO}_3^{-}\) ions, with nitrate ions contributing to a higher overall concentration due to their presence from both lead nitrate and chromic nitrate sources.